Problem: Drazil likes heap very much. So he created a problem with heap:

There is a max heap with a height h implemented on the array. The details of this heap are the following:

This heap contains exactly 2^h - 1 distinct positive non-zero integers. All integers are distinct. These numbers are stored in the array a indexed from 1 to 2^h-1. For any 1 < i < 2^h, a[i] < a[\left ⌊{i/2}\right ⌋].

Now we want to reduce the height of this heap such that the height becomes g with exactly 2^g-1 numbers in heap. To reduce the height, we should perform the following action 2^h-2^g times:

Choose an index i, which contains an element and call the following function f in index i:

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Note that we suppose that if a[i]=0, then index i don't contain an element.

After all operations, the remaining 2^g-1 element must be located in indices from 1 to 2^g-1. Now Drazil wonders what's the minimum possible sum of the remaining 2^g-1 elements. Please find this sum and find a sequence of the function calls to achieve this value.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 70 000): the number of test cases.

Each test case contain two lines. The first line contains two integers h and g (1 ≤ g < h ≤ 20). The second line contains n = 2^h-1 distinct positive integers a[1], a[2], …, a[n] (1 ≤ a[i] < 2^{20}). For all i from 2 to 2^h - 1, a[i] < a[\left ⌊{i/2}\right ⌋].

The total sum of n is less than 2^{20}.

Output

For each test case, print two lines.

The first line should contain one integer denoting the minimum sum after reducing the height of heap to g. The second line should contain 2^h - 2^g integers v_1, v_2, …, v_{2^h-2^g}. In i-th operation f(v_i) should be called.

Example

Input


2
3 2
7 6 3 5 4 2 1
3 2
7 6 5 4 3 2 1


Output


10
3 2 3 1
8
2 1 3 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
const double eps = 1e-10;
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp max(const _Tp &a, const _Tp &b) {
  return a < b ? b : a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp min(const _Tp &a, const _Tp &b) {
  return a < b ? a : b;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) bool _cmp(const _Tp &a,
                                                    const _Tp &b) {
  return abs(a - b) <= eps;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void read(_Tp &x) {
  register char ch(getchar());
  bool f(false);
  while (ch < 48 || ch > 57) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (ch >= 48 && ch <= 57)
    x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
__inline__ __attribute__((always_inline)) int read_str(char *s) {
  register char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  register char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = (1 << 21) + 5;
int a[N];
bool tag[N];
struct heap {
  std::priority_queue<int> q1, q2;
  __inline__ __attribute__((always_inline)) void push(int o) { q1.push(o); }
  __inline__ __attribute__((always_inline)) void pop(int o) { q2.push(o); }
  __inline__ __attribute__((always_inline)) int top() {
    while (!q1.empty() && !q2.empty() && q1.top() == q2.top())
      q1.pop(), q2.pop();
    return q1.top();
  }
  __inline__ __attribute__((always_inline)) void clear() {
    while (!q1.empty()) q1.pop();
    while (!q2.empty()) q2.pop();
  }
} q;
int g;
int nd;
int id[N];
void del(int i) {
  int l = i << 1;
  int r = i << 1 | 1;
  if (!a[l] && !a[r]) {
    a[i] = 0;
    tag[i] = false;
    nd = i;
  } else {
    bool last;
    if (a[l] > a[r]) {
      a[i] = a[l];
      last = tag[l];
      del(l);
    } else {
      a[i] = a[r];
      last = tag[r];
      del(r);
    }
    if (!a[i << 1] && !a[i << 1 | 1]) {
      tag[i] = i >= (1 << g);
    } else {
      tag[i] = tag[a[i << 1] > a[i << 1 | 1] ? (i << 1) : (i << 1 | 1)];
    }
    if (last && !tag[i]) q.pop(a[i]);
    id[a[i]] = i;
  }
}
void MAIN() {
  q.clear();
  int h;
  read(h, g);
  memset(a, 0, (1 << h) << 3);
  memset(tag, 0, (1 << h) << 1);
  long long sum = 0;
  for (int i = 1; i < 1 << h; ++i) read(a[i]), sum += a[i];
  int n = (1 << h) - 1;
  for (int i = 1; i <= n; ++i) {
    tag[i] = true;
    q.push(a[i]);
    id[a[i]] = i;
  }
  std::vector<int> v;
  for (int T = 1, lim = (1 << h) - (1 << g); T <= lim; ++T) {
    int x = q.top();
    q.pop(x);
    v.push_back(id[x]);
    del(id[x]);
    int i = id[x] >> 1;
    while (i) {
      if (tag[i]) {
        if (!a[i << 1] && !a[i << 1 | 1]) {
          tag[i] = i >= (1 << g);
        } else {
          tag[i] = tag[a[i << 1] > a[i << 1 | 1] ? (i << 1) : (i << 1 | 1)];
        }
        if (!tag[i]) {
          q.pop(a[i]);
        }
      }
      i >>= 1;
    }
    sum -= x;
    id[x] = 0;
  }
  printf("%lld\n", sum);
  for (auto it : v) printf("%d ", it);
  printf("\n");
}
int main() {
  int _;
  read(_);
  while (_--) MAIN();
  return 0;
}